# GATE EC 2005

 Question 1
The following differential equation has
$3(\frac{d^{2}y}{dt^{2}})+4(\frac{dy}{dt})^{3}+y^{2}+2=x$
 A degree = 2, order = 1 B degree = 3, order = 2 C degree = 4, order = 3 D degree = 2, order = 3
Engineering Mathematics   Differential Equations
Question 1 Explanation:
Order is highest derivative term. Degree is power of highest derivative term.
 Question 2
Choose the function $f (t); -\infty \lt t \lt \infty$ for which a Fourier series cannot be defined.
 A 3sin(25t) B 4cos(20t+3)+2sin(10t) C exp(-|t|)sin(25t) D 1
Signals and Systems   Fourier Series
Question 2 Explanation:
All other functions are either periodic or constant function.
 Question 3
A fair dice is rolled twice. The probability that an odd number will follow an even number is
 A $1/2$ B $1/6$ C $1/3$ D $1/4$
Engineering Mathematics   Probability and Statistics
Question 3 Explanation:
\begin{aligned} P_{0}=\frac{3}{6}=\frac{1}{2} \\ P_{e}=\frac{3}{6}=\frac{1}{2} \end{aligned}
since both events are independent of each other.
$P_{\text {(odd/even) }}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$
 Question 4
A solution of the following differential equation is given by $\frac{d^{2}y}{dx^{2}}-5\frac{dy}{dx}+6y=0$
 A $y=e^{2x}+e^{-3x}$ B $y=e^{2x}+e^{3x}$ C $y=e^{-2x}+e^{3x}$ D $y=e^{-2x}+e^{-3x}$
Engineering Mathematics   Differential Equations
Question 4 Explanation:
\begin{aligned} A E \Rightarrow\quad D^{2}-5 D+6&=0 \\ (D-2)(D-3) &=0 \\ D &=2,3 \\ \therefore \quad y &=e^{2 x}+e^{3 x} \end{aligned}
 Question 5
The function x(t) is shown in the figure. Even and odd parts of a unit step function u(t) are respectively,
 A $\frac{1}{2},\frac{1}{2}x(t)$ B $-\frac{1}{2},\frac{1}{2}x(t)$ C $\frac{1}{2},-\frac{1}{2}x(t)$ D $-\frac{1}{2},-\frac{1}{2}x(t)$
Signals and Systems   Basics of Signals and Systems
Question 5 Explanation:
$\begin{array}{l} \text { Even part }=\frac{u(t)+u(-t)}{2}\\ \text { Odd part }=\frac{u(t)-u(-t)}{2} \end{array}$

 Question 6
The region of convergence of z-transform of the sequence $(\frac{5}{6})^{n}u(n)-(\frac{6}{5})^{n}u(-n-1)$ must be
 A $|z| \lt \frac{5}{6}$ B $|z| \gt \frac{5}{6}$ C $\frac{5}{6} \lt |z| \lt \frac{6}{5}$ D $\frac{5}{6} \lt |z| \lt \infty$
Signals and Systems   Z-Transform
Question 6 Explanation:
$\begin{array}{l} z \text {-transform of }\left(\frac{5}{6}\right)^{n} u(n)-\left(\frac{6}{5}\right)^{n} u(-n-1)\\ =\frac{1}{1-\frac{5}{6} z^{-1}}+\left[1-\frac{1}{1-\left(\frac{6}{5}\right)^{-1} z}\right] \\ \begin{array}{cc} (\mathrm{ROC}) &\quad(\mathrm{ROC})\\ =|z| \gt \frac{5}{6}&\quad |z| \lt \frac{6}{5} \\ \end{array}\\ \frac{5}{6}\lt |z| \lt \frac{6}{5} \end{array}$
 Question 7
The condition on R, L and C such that the step response y(t) in the figure has no oscillations, is
 A $R\geq \frac{1}{2}\sqrt{\frac{L}{C}}$ B $R\geq \sqrt{\frac{L}{C}}$ C $R\geq 2\sqrt{\frac{L}{C}}$ D $R=\frac{1}{\sqrt{LC}}$
Network Theory   Sinusoidal Steady State Analysis
Question 7 Explanation:
Transfer function
\begin{aligned} =& \frac{\frac{1}{s C}}{R+s L+\frac{1}{s C}}=\frac{1}{s^{2} L C+s C R+1} \\ \frac{Y(s)}{U(s)}=& \frac{\frac{1}{L C}}{s^{2}+\frac{R}{L} s+\frac{1}{L C}} \\ 2 \xi \omega_{n} &=\frac{R}{L} \\ \omega_{n} &=\frac{1}{\sqrt{L C}} \\ \xi &=\frac{R}{2 L} \sqrt{L C}=\frac{R}{2} \sqrt{\frac{C}{L}} \end{aligned}
For no oscillations, $\xi \geq 1$
$\frac{R}{2} \sqrt{\frac{C}{L}} \geq 1 \quad ; \quad R \geq 2 \sqrt{\frac{L}{C}}$
 Question 8
The ABCD parameters of an ideal n:1 transformer shown in the figure are $\begin{bmatrix} n & 0\\ 0&x \end{bmatrix}$

The value of x will be
 A n B $\frac{1}{n}$ C $n^{2}$ D $\frac{1}{n^{2}}$
Network Theory   Two Port Networks
Question 8 Explanation:
\begin{aligned} \left[\begin{array}{l} V_{1} \\ I_{1} \end{array}\right] &=\left[\begin{array}{ll} A & B \\ C & D \end{array}\right]\left[\begin{array}{l} V_{2} \\ -I_{2} \end{array}\right] \\ -\frac{I_{2}}{I_{1}} &=\frac{V_{1}}{V_{2}}=\frac{n}{1} \\ V_{1} &=A V_{2}-B I_{2}\\ I_{1}&=C V_{2}-D I_{2} \\ A&=\left.\frac{V_{1}}{V_{2}}\right|_{I_{2}=0}=n \\ D&=-\left.\frac{I_{1}}{I_{2}}\right|_{V_{2}=0}=\frac{V_{2}}{V_{1}}=\frac{1}{n} \end{aligned}
 Question 9
In a series RLC circuit, R=2k$\Omega$, L=1H, and $C=\frac{1}{400}\mu F$ The resonant frequency is
 A $2 \times 10^{4}Hz$ B $\frac{1}{\pi }\times 10^{4}Hz$ C $10^{4}Hz$ D $2\pi \times 10^{4}Hz$
Network Theory   Sinusoidal Steady State Analysis
Question 9 Explanation:
\begin{aligned} f_{0} &=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{1 \times \frac{1}{400} \times 10^{-6}}} \\ &=\frac{10^{3} \times 20}{2 \pi}=\frac{10^{4}}{\pi} \mathrm{Hz} \end{aligned}
 Question 10
The maximum power that can be transferred to the load resistor $R_{L}$ from the voltage source in the figure is
 A 1 W B 10 W C 0.25 W D 0.5 W
Network Theory   Network Theorems
Question 10 Explanation:
For maximum power transfer
\begin{aligned} R_{L}&= R_{s}=100 \Omega \\ \therefore \quad P_{\max }&=\frac{V^{2}}{R}=\frac{5 \times 5}{100}=0.25 \mathrm{W} \\ &\qquad\left(V \operatorname{across} R_{L}=5 \mathrm{V}\right) \end{aligned}
There are 10 questions to complete.